Another Vector Question

A 2.5 kg block slides down a 25 degree inclined plane with constant acceleration. The block starts from rest at the top. At the bottom, its velocity reaches 0.65 m/s. The length of the inclined plane is 1.6 m. What is the coefficient of friction between the inclined plane and block?

Here is what I did.

By using acceleration and average velocity equations I found that the acceleration of the block is 0.13 m/s/s.

Then friction force = mu times normal force. The weight of the block is 2.5(9.8) = 24.5 N. So the normal force is 24.5cos25 = 22.2.

The force acting on the block parallel to the plane is 24.5sin25 = 10.4.

So net force = 0.13(2.5) = 0.33.

Therefore 0.33 = 10.4 + friction force. Which means friction force = -10.1.

So -10.1 = mu(22.2) which means mu = -0.455.

Is my line of thinking at all correct? Can you have a negative coefficient of friction?

Your problem is here and there is two ways to handle it.

The first is to realize that the frictional force always opposes the movement so you should have done:

0.33 = 10.4 - friction force.

The second is to use 0.33 = 10.4 + friction force

and realize that the equation Friction force = mu times normal force gives you only the magnitude of the frictional force not the direction so you have to drop the negative sign so that:

10.1 = mu*(22.2) which gives you the same answer but without the negative sign. Other than that all the work you did is fine. You just have to remember to keep track of which equations take into account magnitude and direction and which equations just take into account magnitude.

Thanks.

So I take it you can't have a negative coefficient of friction.

Yeah that wouldn't make any sense.